Smallest Difference（剪枝）

最新推荐文章于 2024-11-21 16:03:17 发布

ShawnVan_Bean

最新推荐文章于 2024-11-21 16:03:17 发布

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分类专栏： dfs

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本文介绍了一种解决从一组升序排列的个位数中选取两个子集以形成两个整数，使得这两个整数差的绝对值最小的问题的算法。通过使用双重深度优先搜索策略并加入剪枝技巧，确保了算法的高效性。

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Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some order. The remaining digits can be written down in some order to form a second integer. Unless the resulting integer is 0, the integer may not start with the digit 0.

For example, if you are given the digits 0, 1, 2, 4, 6 and 7, you can write the pair of integers 10 and 2467. Of course, there are many ways to form such pairs of integers: 210 and 764, 204 and 176, etc. The absolute value of the difference between the integers in the last pair is 28, and it turns out that no other pair formed by the rules above can achieve a smaller difference.

Input

The first line of input contains the number of cases to follow. For each case, there is one line of input containing at least two but no more than 10 decimal digits. (The decimal digits are 0, 1, ..., 9.) No digit appears more than once in one line of the input. The digits will appear in increasing order, separated by exactly one blank space.

Output

For each test case, write on a single line the smallest absolute difference of two integers that can be written from the given digits as described by the rules above.

Sample Input

1
0 1 2 4 6 7

Sample Output

题意：给出一个集合，集合里是按升序排列的个位数，将它们划分为两个整数a与b，求a与b差值最小的值
题解：如果总共有n个数，必然有一个整数长n/2,另一个长n-n/2。此题我用双重dfs求解，双dfs 在 a 已定的情况下找 b 找到最小值。其中有一个重要的过程要剪枝，就是当a 一定时，若此时未求出完整的b，但此时b 之后的位都补0 与a 的差大于等于此时的ans（绝对值）则返回。

#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdio.h>
#define INF 0x3f3f3f3f
using namespace std;
int num[25];
int len;
int ans;
int val;
int lena;
int lenb;
bool visa[25];
bool visb[25];
int nums[10]={1,10,100,1000,10000,100000,1000000};
void dfs_b(int b,int deep){
	if(deep>0&&abs(val-b*nums[lenb-deep])>=ans)
		return;
	if(deep==lenb){
		ans=min(ans,abs(val-b));
		return;
	}
	for(int i=0;i<len;i++)
		if(!visa[i]&&!visb[i]){
			if(deep==0&&num[i]==0)
				continue;
			visb[i]=1;
			dfs_b(b*10+num[i],deep+1);
			visb[i]=0;
		}
}
void dfs_a(int a,int deep){
	if(deep==lena){
		val=a;
		memset(visb,0,sizeof(visb));
		dfs_b(0,0);
		return;
	}
	for(int i=0;i<len;i++)
		if(!visa[i]){
			if(deep==0&&num[i]==0)
				continue;
			visa[i]=1;
			dfs_a(a*10+num[i],deep+1);
			visa[i]=0;
		}
}
int main(){
	int t;
	while(cin>>t){
		getchar();
		while(t--){
			char ch;
			len=0;
			while((ch=getchar())!='\n')
				if(ch!=' ')
					num[len++]=ch-'0';
			lena=len/2;
			lenb=len-lena;
			ans=INF;
			dfs_a(0,0);
			if(ans==INF)
				cout<<val<<endl;
			else
				cout<<ans<<endl;
		}
	}
	return 0;
}